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I have an example of an unmeasurable set in the measure space defined as follows: If $\Omega$ is the real line, we can consider a $\sigma$-algebra consisting of all the closed intervals of the real line and a measure that is just the length of the interval. This $\sigma$-algebra does not contain all the subsets of the real line and thus some sets are not measurable in this measure space.

I am confused as to, what are the subsets that are not contained in this measure space? I am guessing the subset of open intervals is missing, but can't all open intervals be expressed as closed intervals?

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  • $\begingroup$ Why do you think the set of all closed intervals of the real line is a $\sigma$-algebra? What is your definition of a $\sigma$-algebra? $\endgroup$ – bof Jan 24 '18 at 22:16
  • $\begingroup$ What makes you think all open intervals can be expressed as closed intervals? How would you express the open interval $(0,1)$ as a closed interval? $\endgroup$ – bof Jan 24 '18 at 22:17
  • $\begingroup$ @bof I think I am bit confused there. The difference between open and closed interval is that the open interval does not contain boundary points. Doesnt that mean we have a new boundary that we can define the open interval with making it a closed interval? $\endgroup$ – Anuroop Kuppam Jan 24 '18 at 22:30
  • $\begingroup$ It looks like you need experience with "open" and "closed" before you advance to $\sigma$-algebras. $\endgroup$ – GEdgar Jan 24 '18 at 23:46
  • $\begingroup$ The set of closed intervals is not a $\sigma$-algebra. Check the def'n. $\endgroup$ – DanielWainfleet Jan 25 '18 at 3:05
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Not everything is a closed interval or a union of closed intervals. For example, $(0,1)$, $\mathbb{Q}$, $\mathbb{R\backslash Q}$, and the Cantor Set. Later on these can assign a measure to these sets, but in the way you defined measure they are non-measurable.

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